I need to prove that the following two definitions of the zero-th $K$-theory group of a ring $R$ (with unit) are equivalent.
def 1 the abelian group generated by the isomorphism classes $[P]$ of $R$-projective modules (finitely generated) under the identification $[P\oplus Q]=[P]+[Q]$
def2 The Grothendieck group of the monoid of isomorphism classes of finitely generated $R$-proj. module, with operation the direct sum.
My attempt Call $K^1$ the abelian group with the first def, and $K^2$ the group with the second one. I defined a map $K^1 \to K^2$ as $$\lambda[P] \mapsto (\oplus_{\lambda}[P],0)$$ and $$-\lambda[Q] \mapsto (0,\oplus_{\lambda}[Q])$$ for $\lambda$ positive integer. Now the map is clearly surjective and well-defined. But I don't know how to prove injectiviness. I mean, should be fairly easy but I got stuck in th following step: for simplicity assume that $[P]-[Q] \mapsto ([P],[Q])\sim (0,0)$. By definition of Grothendieck group there exists a finely generated proj module $C$ such that $P\oplus C \simeq Q \oplus C$. But now I don't know how to get rid of such $C$ in order to prove that $0=[P]-[Q]$. Any ideas?
If $P \oplus C \cong Q \oplus C$, then $[P \oplus C] = [Q \oplus C]$ in $K^1$, hence $[P] + [C] = [Q] + [C]$ in $K^1$, hence $[P] - [Q] = 0$ in $K^1$.